The coefficients of three successive terms in the expansion of (1+x)n  are 165, 330 and 462 respectively, then the value of n  will be

Given expansion  (1+x)n is We have general term in the expansion (x+a)n (∴  Tr+1= nCr xn−r (a)r  be the expansion of (x+a)n) Let the three successive terms be Tr,Tr+1  and Tr+2 , then rth  term:  T(r−1)+1= nCr−1 xn−r+1 (a)r−1  Coefficient of rth  term is  nCr−1 ∵ nCr−1=165..................(1)       (r+1)th  term:  Tr+1= nCr xn−r (a)r  Coefficient of (r+1)th  term is  nCr ∵ nCr=330..................(2) (r+2)th  term: T(r+1)+1= nCr+1 xn−r−1 (a)r+1   Coefficient of (r+2)th  term is  nCr+1 ∵nCr+1=462.........................(3) Dividing Eqn (2)by Eqn (1), we get nCrnCr−1=330165 (∴nCrnCr−1=n−r+1r)           n−r+1r=2            …(4) Dividing Eqn (3)by Eqn (2), we get nCr+1nCr=462330 (∴nCrnCr−1=n−r+1r)          n−rr+1=75                  ….(5)From (4), n=3r−1  and from (5)                           n=  12r+75 ⇒  3r−1=12r+75  ⇒  3r=12 ∵r=4  in  Eqn (5) we get n=11